# Numbers with multiple runs of ones

Find the set of numbers such that the binary representation contains two or more runs of 1 separated by at least one 0.

For example, the for the numbers that are 4 bits long:

 0 0000        (no ones)
1 0001        (only one run)
2 0010        (only one run)
3 0011        (only one run)
4 0100        (only one run)
5 0101 Valid
6 0110        (only one run)
7 0111        (only one run)
8 1000        (only one run)
9 1001 Valid
10 1010 Valid
11 1011 Valid
12 1100        (only one run)
13 1101 Valid
14 1110        (only one run)
15 1111        (only one run)


### Input

An integer provided to the application via some input in the range 3 .. 32. This represents the maximum number of bits to count up to.

The input of n indicates that the numbers 0 .. 2n-1 need to be examined.

### Output

A delimited (your choice) list of all numbers meeting the criteria. The numbers are to be presented in numeric order. An extra trailing delimiter is acceptable. Data structure enclosures (e.g. [] and similar) are also acceptable.

### Example

Input: 3
Output: 5

Input: 4
Output: 5, 9, 10, 11, 13

Input: 5
Output: 5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29


This is - the answer with the least amount of bytes wins.

• I think you missed 23 for n=5.
– xnor
Oct 21, 2015 at 21:02
• @xnor you are correct. Thank you, and yep, that also makes it not equivalent to A094695. Hmm. oeis.org/A101082 vs oeis.org/A166934
– user12166
Oct 21, 2015 at 21:03
• @VTCAKAVSMoACE yes. If one is \n delimiting and putting a \n on the last line, then , delimited with a , trailing should be acceptable too. Updated.
– user12166
Oct 21, 2015 at 21:24
• Can the input be in a list format like [1, 2, 3]? Oct 21, 2015 at 22:33
• @kirbyfan64sos yes. Updated.
– user12166
Oct 21, 2015 at 23:11

## Python, 48

lambda n:[i for i in range(2**n)if'01'in bin(i)]


I had been vastly overthinking this. We just need to check if the binary expansion contains '01'.

For there to be two runs of ones, the one on the right must be preceded by a 0. If there's only one run, there won't be any leading 0's, so that won't happen.

lambda n:[i for i in range(2**n)if len(set(bin(i).split('0')))>2]


The Python binary representation works very nicely here. A binary number is written like bin(9)=='0b10110'. Splitting at '0' results in a list of

• Empty strings to the left of the initial 0, between any two consecutive 0's, and to the right of any final 0
• The letter b followed by one or more leading ones
• Runs of 1's that are not leading

The first two categories always exist, but the last one only exists if there is a run one 1's that doesn't contain the leading '1', and so only if there's more than one run of 1's. So, it suffices to check if the list contains more than 2 distinct elements.

Python 3.5 saves 2 chars by unpacking {*_} in place of set(_).

• Thanks for the idea to use /01/ instead of /10+1/. I took advantage of that in Perl. Oct 21, 2015 at 22:48

## Ruby, 4440 38 chars

crossed out 44 is still regular 44 ;(

->n{(0..2**n).select{|x|/01/=~'%b'%x}}


An anonymous function (proc, actually) that takes an integer and returns an array.

Uses the regex /10+1/: a 1, at least one 0, and then another 1. @histocrat points out that if 01 is anywhere in the string, there must be a 1 somewhere before it.

• Using a format string is a bit shorter here: /10+1/=~'%b'%x. Also, you can save a character by using an inclusive range(0..2**n) since 2**n will never have multiple runs. Oct 21, 2015 at 21:07
• @histocrat Huh, I never knew you could flip the order of the string and the regex with =~. Thanks! Oct 21, 2015 at 21:10
• Wait, actually the regex /01/ works just as well. If there's a 01, there must be a 1 to the left somewhere. Oct 21, 2015 at 21:27
• @histocrat Oh, that's clever! That saves two characters. Oct 21, 2015 at 21:30

# Pyth, 12 bytes

f<2r.BT8U^2Q


Try it online.

### Idea

The binary representation of any positive number always begins with a run of 1s, possibly followed by other, alternating runs of 0s and 1s. If there are at least three separate runs, two of them are guaranteed to be runs of 1s.

### Code

              (implicit) Store the evaluated input in Q.
^2Q  Calculate 2**Q.
f       U     Filter; for each T in [0, ..., 2**Q-1]:
.BT         Compute T's binary representation.
r   8        Perform run-length encoding.
This returns a list of character/run-length pairs.
<2             Discard the trailing two pairs.
This returns a non-empty array if there are more than 2 runs.
Keep T if the array was truthy (non-empty).


# Julia, 43 41 bytes

n->filter(i->ismatch(r"01",bin(i)),1:2^n)


This creates an unnamed function that accepts an integer and returns an array. It uses histocrats's regex trick (used in Doorknob's answer), where 01 will only match if there's a preceding 1.

Ungolfed:

function f(n::Int)
# Take the integers from 1 to 2^n and filter them down to
# only those such that the binary representation of the integer
# matches the regex /01/.
filter(i -> ismatch(r"01", bin(i)), 1:2^n)
end

• histocrat's trick, not mine. :) Oct 22, 2015 at 20:48
• @Doorknob Oh hey, now you both get credit. :) Oct 22, 2015 at 20:49

# Matlab, 79 68 64 59

The idea is interpreting the binary number as array of zeros and ones, and then calculating the absolute difference between each pair of neighbours. If we have two or more times a difference of 1, then we obviously have a run of two or more ones. Note that this only works if we represent the binary number without leading zeros.

@(n)find(arrayfun(@(k)sum(~~diff(dec2bin(k)+0))>1,1:2^n-1))


Old versions:

k=1:2^input('')-1;k(arrayfun(@(k)sum(~~diff(dec2bin(k)+0))>1,k))

for k=1:2^input('')-1;if sum(~~diff(dec2bin(k)+0))>1;disp(k);end;end

for k=1:2^input('')-1;if sum(~~conv(dec2bin(k)+0,[-1,1],'v'))>1;disp(k);end;end


# JavaScript (ES7), 89857269 62 bytes

Holy cow, creating ranges in JS is not easy. Perhaps it would be shorter with an actual for loop. Nope, I lied; it's actually a bit longer. Oh well. I guess I'll just have to settle for 27 bytes saved. (7 thanks to Mwr247!)

x=>[for(a of Array(1<<x).keys())if(/01/.test(a.toString(2)))a]


Works properly in the latest versions of Firefox, but probably not in any other browser. Try it out:

<!--                               Try the test suite below!                              --><strong id="bytecount" style="display:inline; font-size:32px; font-family:Helvetica"></strong><strong id="bytediff" style="display:inline; margin-left:10px; font-size:32px; font-family:Helvetica; color:lightgray"></strong><br><br><pre style="margin:0">Code:</pre><textarea id="textbox" style="margin-top:5px; margin-bottom:5px"></textarea><br><pre style="margin:0">Input:</pre><textarea id="inputbox" style="margin-top:5px; margin-bottom:5px">5</textarea><br><button id="testbtn">Test!</button><button id="resetbtn">Reset</button><br><p><strong id="origheader" style="font-family:Helvetica; display:none">Original Code Output:</strong><p><div id="origoutput" style="margin-left:15px"></div><p><strong id="newheader" style="font-family:Helvetica; display:none">New Code Output:</strong><p><div id="newoutput" style="margin-left:15px"></div><script type="text/javascript" id="golfsnippet">var bytecount=document.getElementById("bytecount");var bytediff=document.getElementById("bytediff");var textbox=document.getElementById("textbox");var inputbox=document.getElementById("inputbox");var testbtn=document.getElementById("testbtn");var resetbtn=document.getElementById("resetbtn");var origheader=document.getElementById("origheader");var newheader=document.getElementById("newheader");var origoutput=document.getElementById("origoutput");var newoutput=document.getElementById("newoutput");textbox.style.width=inputbox.style.width=window.innerWidth-50+"px";var _originalCode="x=>[for(a of Array(1<<x).keys())if(/01/.test(a.toString(2)))a]";function getOriginalCode(){if(_originalCode!=null)return _originalCode;var allScripts=document.getElementsByTagName("script");for(var i=0;i<allScripts.length;i++){var script=allScripts[i];if(script.id!="golfsnippet"){originalCode=script.textContent.trim();return originalCode}}}function getNewCode(){return textbox.value.trim()}function getInput(){try{var inputText=inputbox.value.trim();var input=eval("["+inputText+"]");return input}catch(e){return null}}function setTextbox(s){textbox.value=s;onTextboxChange()}function setOutput(output,s){output.innerHTML=s}function addOutput(output,data){output.innerHTML+='<pre style="background-color:'+(data.type=="err"?"lightcoral":"lightgray")+'">'+escape(data.content)+"</pre>"}function getByteCount(s){return(new Blob([s],{encoding:"UTF-8",type:"text/plain;charset=UTF-8"})).size}function onTextboxChange(){var newLength=getByteCount(getNewCode());var oldLength=getByteCount(getOriginalCode());bytecount.innerHTML=newLength+" bytes";var diff=newLength-oldLength;if(diff>0){bytediff.innerHTML="(+"+diff+")";bytediff.style.color="lightcoral"}else if(diff<0){bytediff.innerHTML="("+diff+")";bytediff.style.color="lightgreen"}else{bytediff.innerHTML="("+diff+")";bytediff.style.color="lightgray"}}function onTestBtn(evt){origheader.style.display="inline";newheader.style.display="inline";setOutput(newoutput,"");setOutput(origoutput,"");var input=getInput();if(input===null){addOutput(origoutput,{type:"err",content:"Input is malformed. Using no input."});addOutput(newoutput,{type:"err",content:"Input is malformed. Using no input."});input=[]}doInterpret(getNewCode(),input,function(data){addOutput(newoutput,data)});doInterpret(getOriginalCode(),input,function(data){addOutput(origoutput,data)});evt.stopPropagation();return false}function onResetBtn(evt){setTextbox(getOriginalCode());origheader.style.display="none";newheader.style.display="none";setOutput(origoutput,"");setOutput(newoutput,"")}function escape(s){return s.toString().replace(/&/g,"&amp;").replace(/</g,"&lt;").replace(/>/g,"&gt;")}window.alert=function(){};window.prompt=function(){};function doInterpret(code,input,cb){var workerCode=interpret.toString()+";function stdout(s){ self.postMessage( {'type': 'out', 'content': s} ); }"+" function stderr(s){ self.postMessage( {'type': 'err', 'content': s} ); }"+" function kill(){ self.close(); }"+" self.addEventListener('message', function(msg){ interpret(msg.data.code, msg.data.input); });";var interpreter=new Worker(URL.createObjectURL(new Blob([workerCode])));interpreter.addEventListener("message",function(msg){cb(msg.data)});interpreter.postMessage({"code":code,"input":input});setTimeout(function(){interpreter.terminate()},1E4)}setTimeout(function(){getOriginalCode();textbox.addEventListener("input",onTextboxChange);testbtn.addEventListener("click",onTestBtn);resetbtn.addEventListener("click",onResetBtn);setTextbox(getOriginalCode())},100);function interpret(code,input){window={};alert=function(s){stdout(s)};window.alert=alert;console.log=alert;prompt=function(s){if(input.length<1)stderr("not enough input");else{var nextInput=input;input=input.slice(1);return nextInput.toString()}};window.prompt=prompt;(function(){try{var evalResult=eval(code);if(typeof evalResult=="function"){var callResult=evalResult.apply(this,input);if(typeof callResult!="undefined")stdout(callResult)}}catch(e){stderr(e.message)}})()};</script>

Suggestions welcome!

• You can use .keys() instead of .fill() and a instead of i to tie mine for 62: x=>[for(a of Array(1<<x).keys())if(/01/.test(a.toString(2)))a] Oct 22, 2015 at 20:55
• @Mwr247 Thanks! I wonder if it's possible in under 62... :) Oct 22, 2015 at 23:30

Improvement from Damien

g x|xmod4==1=x>4|2>1=g$xdiv2 a x=filter g[1..2^x]  History: This fixes the bug(Switched == and =, and square instead of power of two). And replace true with 2>1 and false with 1>2. Also thanks to point out that 2^x is always fail. Thanks to Thomas Kwa and nimi g x|x<5=1>2|xmod4==1=2>1|2>1=g$xdiv2
a x=filter g[1..2^x]


Originally

g x|x<5=False|xmod4=1==True|2>1=g$xdiv2 a x=filter g[1..(x^2-1)]  If it have to be full program, g x|x<5=False|xmod4==1=True|2>1=g$xdiv2
main=interact$show.a a x=filter g[1..2^(read x)]  • Lambdas are fine, since OP didn't specify writing a named function or program. By the way, welcome to PPCG! Oct 22, 2015 at 15:29 • I think you mean 1..(2^x-1) which can just be 1.. (2^x) since 2^x always fails. Oct 22, 2015 at 16:45 • You can replace the constants False and True with 1>2 and 1<2. No need for the parentheses around 2^x-1. (BTW: you have a typo: it must be 4==1=True). – nimi Oct 22, 2015 at 17:56 • Thanks for typo correction. It was late night in my time. Oct 23, 2015 at 5:29 • nice tricks!I think you can reduce g to: g x|xmod4==1=x>4|2>1=g$xdiv2 Oct 23, 2015 at 11:47

# APL, 34 27 bytes

{0~⍨{⍵×2<+/2≢/⍵⊤⍨⍵/2}¨⍳2*⍵}


This creates an unnamed monadic function that accepts an integer on the right and returns an array.

Explanation:

                     }¨⍳2*⍵}  ⍝ For each integer from 1 to 2^input...
⍵⊤⍨⍵/2         ⍝ Get the binary representation as a vector
2≢/                ⍝ Pairwise non-match, yielding a boolean vector
2<+/                   ⍝ Check whether the number of trues is >2
⍵×                       ⍝ Yield the integer if so, otherwise 0
{0~⍨{                         ⍝ Remove the zeros from the resulting array


Saved 7 bytes thanks to Dennis!

# C (gcc), 111 99 bytes

long i,x;main(a,b)char**b;{for(;++i<1L<<atol(b);x>>ffsl(~x)-1&&printf("%ld,",i))x=i>>ffsl(i)-1;}


Try it online!

12 bytes shaved of thanks to @ceilingcat!

Ungolfed:

int main(int a, char **b) {
for(long i = 0, x = 0; ++i < (1LL << atol(b)); ) {
x = i >> (ffsl(i) - 1);
if (x >> (ffsl(~x) - 1))
printf("%ld,", i);
}
}


The function ffsl() gives you the index of the first bit that is set in a long integer. So we loop from i = 1 to 2^number_of_bits. We set x to i shifted right until we have removed all consecutive zero bits on the least significant end. Then, we shift x right until we have removed all consecutive 1 bits at the least signficant end. If the result is still non-zero, we found a match.

• I've gotta say that I really like that someone did a bit manipulation answer rather than a "convert to string and regex" approach.
– user12166
Oct 22, 2015 at 1:45
• @MichaelT I wonder if there's a short olution using only primitive bitwise operations. Oct 22, 2015 at 1:50
• @ThomasKwa That might be something to do as a code-challenge.
– user12166
Oct 22, 2015 at 1:54
• Interesting. You can also write the test like this: if (popcount(i ^ (i*2))>3), and expand popcount() to a series of bitwise ANDs and shift operations. But that would result in quite long code. Oct 22, 2015 at 6:57
• @ThomasKwa y=x|(x-1) to turn on all rightmost 0 bits. Then z=y&(y+1) to turn off all trailing 1 bits. If z is non-zero, then the original number had more than one run. Oct 26, 2015 at 14:41

# R, 55 47 bytes

(with some help from @Alex.A)

cat(grep("10+1",R.utils::intToBin(1:2^scan())))


R doesn't have a built in function to display converted numbers in a convenient way, so I'm using R.utils::intToBin for this, while all the rest is pretty much just report the location of the matched regex expression and print to STDOUT while separated by a space.

• I think the default separator for cat is a space, so you could omit ,sep="," entirely, saving 7 bytes. Oct 21, 2015 at 22:45
• @AlexA. yeah, so can I use a space here as a sep? I wasn't sure Oct 21, 2015 at 22:47
• The OP said a delimiter of your choice, so I think a space seems reasonable enough. :) Oct 21, 2015 at 22:48
• Does this really need the cat function? without it, it the output will be tab delimited. The left hand counter is part of the UI, if you write it to a file this won't be included so it's not part of the output. Oct 23, 2015 at 12:02
• @freekvd without it won't print to STDOUT, Something about the silly rules of this site. Oct 23, 2015 at 12:18

# CJam, 14

2qi#{2b2,#)},p


3 bytes shorter thanks to Dennis. Try it online

• How about 2be,2>. Oct 21, 2015 at 23:27
• 2be2> and 2,#) should work as well. Also, the OP has clarified that the output can be printed in list form. Oct 22, 2015 at 0:06

# JavaScript (ES6), 696867 62 bytes

a=>[...Array(1<<a).keys()].filter(i=>/01/.test(i.toString(2)))


Today I discovered a new shorter way to dynamically fill arrays without the use of fill or map. Doing x=>[...Array(x).keys()] will return an array of range 0 to x. If you want to define your own range/values, use x=>[...Array(x)].map((a,i)=>i), as it's just a few bytes longer.

## Java, 214165155154148141 110 bytes

This submission exploits the fact that a binary string representation of a number in Java never has a leading zero. If the string "01" appears in the binary representation of a number, that must mark the second occurrence of the number "1".

Golfed:

String f(int l){String r="";for(long i=5;i<1L<<l;++i)if(Long.toString(i,2).contains("01"))r+=i+", ";return r;}


Ungolfed:

public class NumbersWithMultipleRunsOfOnes {

public static void main(String[] a) {
// @formatter:off
String[][] testData = new String[][] {
{ "3", "5" },
{ "4", "5, 9, 10, 11, 13" },
{ "5", "5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29" }
};
// @formatter:on

for (String[] data : testData) {
System.out.println("Input: " + data);
System.out.println("Expected: " + data);
System.out.print("Actual:   ");
System.out.println(new NumbersWithMultipleRunsOfOnes().f(Integer.parseInt(data)));
System.out.println();
}
}

// Begin golf
String f(int l) {
String r = "";
for (long i = 5; i < 1L << l; ++i)
if (Long.toString(i, 2).contains("01")) r += i + ", ";
return r;
}
// End golf
}


Program output (remember, trailing delimiters are acceptable):

Input: 3
Expected: 5
Actual:   5,

Input: 4
Expected: 5, 9, 10, 11, 13
Actual:   5, 9, 10, 11, 13,

Input: 5
Expected: 5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29
Actual:   5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29,

• Couln't you use int for the counter variable? Oct 21, 2015 at 21:23
• All integer types in Java are unsigned. In order to work with a 32 bit positive integer, a 64 bit long is required. Furthermore, using an int would actually increase the size of the code due to referencing the Integer wrapper class which does the number parsing. I think the likely place to save space would be the regex, but my testing showed I have to have leading and trailing .*
– user18932
Oct 21, 2015 at 21:28
• Oh right, but I thought you could use the Long wrapper with int? (Well not in this case but generally?) Oct 21, 2015 at 21:34
• Yes, the int will promote to long when used as a parameter with Long. In this case though there is not really any way to use an int because of the sign bit, and Integer is longer than Long. Still, I have found a few ways to squeeze some extra space out of a language as verbose as Java.
– user18932
Oct 21, 2015 at 21:38
• Can you use new Long() instead of Long.parseLong()? Oct 21, 2015 at 23:57

# JavaScript (ES6) 76

f=n=>Array(1<<n).fill().map((_,x)=>/01/.test(x.toString(2))?x+',':'').join

//TEST
for(i=1;i<16;i++)O.innerHTML+=i+' -> '+f(i)+'\n'
<pre id=O></pre>

• @DLosc no, the result would be something like ,,,,,5,,,,9,10,11,,13,,,,17,18,19,20,21,22,23,,25,26,27,,29,, Oct 22, 2015 at 7:45

# K5, 19 bytes

This operates along similar principles as Dennis' solution, but with fewer builtins to take advantage of.

{&2<+/'~0=':'+!x#2}


First, generate a series of binary x-tuples (+!x#2), then for each tuple find every point that a digit does not match the previous if we treat the -1st element of the list as 0 for this purpose (~0=':'). Our solutions are where two is less than the sum of each run count. (&2<+/').

Showing each intermediate step is clearer:

  4#2
2 2 2 2

!4#2
(0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1)

+!4#2
(0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1)

~0=':'+!4#2
(0 0 0 0
0 0 0 1
0 0 1 1
0 0 1 0
0 1 1 0
0 1 1 1
0 1 0 1
0 1 0 0
1 1 0 0
1 1 0 1
1 1 1 1
1 1 1 0
1 0 1 0
1 0 1 1
1 0 0 1
1 0 0 0)

+/'~0=':'+!4#2
0 1 2 1 2 3 2 1 2 3 4 3 2 3 2 1

2<+/'~0=':'+!4#2
0 0 0 0 0 1 0 0 0 1 1 1 0 1 0 0

&2<+/'~0=':'+!4#2
5 9 10 11 13


And all together:

  {&2<+/'~0=':'+!x#2}'3 4 5
(,5
5 9 10 11 13
5 9 10 11 13 17 18 19 20 21 22 23 25 26 27 29)


# Pip, 13 + 1 = 14 bytes

Takes input from the command line and uses the -s flag for spaces between output numbers.

01NTB_FI,2**a


Pretty straightforward: build range(2**a) and filter on lambda _: "01" in toBinary(_). I was pretty happy about thinking up the 01 idea independently. No quotes are needed around 01 because it scans as a numeric literal (numbers and strings are the same type in Pip).

# Julia, 40 bytes

n->filter(i->count_ones(i$i>>1)>2,1:2^n)  This uses a somewhat different approach to the other Julia solution - rather than doing a string search for "01" in the bit string, it uses some mathematics to determine whether the number satisfies the condition. i$i>>1 will have ones only in the places where the digit changes from zero to one, or one to zero. As such, there must be at least three ones for i to switch back and forth between zero and one enough times. count_ones finds the number of ones, and then filter removes the ones that don't have enough ones.

## C++, 197188 141 bytes

Note: this was written and tested using MSVC++ 2013. It appears that #includeing <iostream> includes all of the necessary C headers to make this work. It also appears that the code is no longer truly C++, but compiling using C++ allows that header trick which reduces the code size compared to including a bunch more C headers.

Using printf instead of cout also saves a couple bytes.

Golfed:

#include<iostream>
int main(int a,char**b){char c;for(long i=5;i<1L<<atol(b);++i){_ltoa(i,c,2);if(strstr(c,"01"))printf("%ld\n",i);}}


Ungolfed:

#include <iostream>
int main(int a, char **b) {
char c;
for (long i = 5; i < 1L << atol(b); ++i) {
_ltoa(i, c, 2);
if (strstr(c, "01"))
printf("%ld\n", i);
}
}

• Yoiu can use '\n' instead of std::endl (in general), or ',' since any separator is valid and a trailing one is fine. Oct 22, 2015 at 0:10
• Instead of regexs, you could just do it with strstr(c,"01"). Oct 22, 2015 at 0:14
• @G.Sliepen thanks! I honestly just copied my Java solution and converted to C++, but the simple solution is often best. I should probably do something similar with Java now.
– user18932
Oct 22, 2015 at 1:32
• Two small errors: 1<<atol(b) should be 1L<<atol(b), otherwise the result of that expression will be a signed 32 bit integer, which means the code will only run up to 2^30. The printf should use %ld to print numbers between 2^31 and 2^32 correctly. Oct 22, 2015 at 6:45

## Perl 5, 55534947 41 bytes

sprintf("%b",$_)=~/01/&&say for 0..2**<>  54 52 48 46 40 bytes, plus one for the -E flag instead of -e. Thanks to xnor for the hint about using /01/ instead of /10+1/, which saved two bytes. Thanks to Dennis for the advice to use <> instead of $ARGV, which saved six bytes.

C, 84 81 bytes

long i,j,k;main(){for(scanf("%ld",&j);++i<1L<<j;k&k+1&&printf("%ld ",i))k=i|i-1;}


This is based on the comments I made on another C answer to this question about the possibility of using simple bitwise operators. It works by switching all trailing 0 bits to 1 in the statement i|(i-1). Then it switches all trailing 1 bits to 0 using k&(k+1). This will result in a zero if there is only one run of ones and non-zero otherwise. I do make the assumption that long is 64-bit but could correct this at the expense of three bytes by using int64_t instead.

Ungolfed

long i,j,k;
main()
{
for(scanf("%ld",&j);++i<1L<<j;)
{
k=i|(i-1);
if((k&(k+1)) == 0)
printf("%d ",i);
}
}

• int64_t is only defined if you #include<stdint.h>. ensuring 64 bit operation requires long long type at a cost of 5 bytes. Oct 29, 2015 at 10:55
• Note that you invoke undefined behaviour passing long i for %d format. Note also that () are superfluous for the & and | operators. Fixing this saves 3 bytes! long i,j,k;main(){for(scanf("%ld",&j);++i<1L<<j;k&k+1&&printf("%ld ",i))k=i|i-1;} Oct 29, 2015 at 10:57
• @chqrlie All very good points. Thank you. Oct 29, 2015 at 16:08

# Pyth, 2017 16 bytes

f:.BT"10+1"ZU^2Q


Live demo.

• You can simply match "01". Except for 0, the binary repr. will always start with a 1. Oct 22, 2015 at 0:08

# Python 2.7, 89 bytes

print[i for i in range(1,2**input())if[n[:1]for n in bin(i)[2:].split("0")].count("1")-1]


I think this could be golfed a bit.

• @mbomb007 I tried that, it didn't work. Oct 21, 2015 at 20:57
• @mbomb007 Are you using Python 2.7? Oct 21, 2015 at 21:03
• Does it matter which version of 2.7? I run it on repl.it (2.7.2) and it doesn't work, but Ideone (2.7.10) does. It might just be a bug in repl.it though, not necessarily a version difference. Oct 21, 2015 at 21:29
• Your program incorrectly prints 0 in the output. Oct 21, 2015 at 21:33
• Also print[i for i in range(2**input())if[n[:1]for n in bin(i)[2:].split("0")].count("1")-1] is two bytes shorter. But with the fix for 0 would be the same length (89), if you use range(1,2**input()). Oct 21, 2015 at 21:35

# TI-BASIC, 3432 30 bytes

For a TI-83+/84+ series calculator.

For(X,5,e^(Ans
If log(sum(2=int(4fPart(X/2^randIntNoRep(1,Ans
Disp X
End


For a number to contain two runs of 1s, it must contain two 10s when we tack a trailing zero onto the binary representation.

Rather than generate the binary representation and check for a 10, we test pairs of bits mathematically by using remainder by 4 (int(4fPart(), which will give 2 where there is a 10. Because we don't care about order, randIntNoRep( is the shortest way to generate the list of exponents.

We use log( to check for the number of runs:

• If there are at least 2 runs, then the log( is positive, and the number is displayed.

• If there is one run, then the log( is 0, and the number is not displayed.

• If there are no runs (which first happens at X=2^Ans), then log( throws an ERR:DOMAIN, stopping the output at exactly the right point.

We use e^(Ans as the ending argument of the For( loop—it's always greater than 2^Ans, but e^( is a single token, so it's one byte shorter.

Input/output for N=4:

4:prgmRUNSONES
5
9
10
11
13


Then the calculator throws an error; the error screen looks like this:

ERR:DOMAIN
1:Quit
2:Goto


When 1 is pressed, the home screen is displayed again:

4:prgmRUNSONES
5
9
10
11
13
Error


TI calculators store all numbers in a BCD float with 14 digits of precision, not an int or binary float. Therefore, divisions by powers of two greater than 2^14 may not be exact. While I have verified that the trickiest numbers, 3*2^30-1 and 2^32-1, are handled correctly, I have not ruled out the possibility of rounding errors. I would however be surprised if there were errors for any input.

• How do you count 32 bytes? It looks like 70 to me (including the newlines). Oct 22, 2015 at 16:46
• TI-BASIC is tokenized; it uses a proprietary character encoding in which all of these commands are one byte each and the others are two. It is the community consensus to score by this encoding-- see meta.codegolf.stackexchange.com/a/4764/39328 for details. Oct 22, 2015 at 16:52
• Oh, cool. Thanks FYI. Oct 22, 2015 at 16:55
• this doesnt beat flawr's answer but i couldnt resist the charm of the question

## matlab(90)(70)

j=input('');for l=2:j-1,a=1;for k=l:-1:2,a=a+2^k;a:a+2^(k-1)-2,end,end


## execution

4

ans =

5


ans =

9    10    11


ans =

13


## principle

• The series of numbers are a result of consequent strip of 1's, which does mean f(n,l)=2^l+2^(l+1)+....2^n

Any number taken from the interval ]f(n,l),f(n,l)+2^(l-1)[ where l>1 verifies this condition, so the outcome is a result of the negation of this series in terms of n.

x=1

x=x+1=01,

x=x+2^0=11,

x=x+1=001,

x=x+2^1=011,

x=x+2^0=111,

x=x+1=0001,

x=x+2^2=0011,

x=x+2^1=0111,

x=x+2^0=0111,

x=x+1=1111 ...

x+1, x=x+2^n, x=x+2^(n-1)...x=x+2^0

My program prints the range between each two lines (if exists)

Edit: unfortunately that doesnt make it golfed more but i wanted to add another approach of proceding this idea

after a period of struggle i succeded to find a mathematical representation for this series which is:

2^l(0+1+2^1+...2^k) with l+k < n

=2^l(2^k-1)

score=90

@(n)setdiff(0:2^n-1,arrayfun(@(x)2^mod(x,(n+1)-fix(x/(n+1)))*(2^fix(x/(n+1))-1),0:(n+1)^2))


# C, 103 102 bytes

long i,x;main(int a,char**b){for(;++i<1L<<atoi(b);)for(x=i;x>4&&(x%4!=1||!printf("%ld,",i));x/=2);}


Expanding (actually contracting) on G.Sliepen entry, taking advantage of xnor remark on the 01 pattern in the binary representation, but using only standard functions and some bit twiddling.

Ungolfed version:

long i, x;
main(int a, char**b) {
for (; ++i < 1L << atoi(b);) {
for (x = i; x > 4 && (x % 4 != 1 || !printf("%ld,", i)); x /= 2)
;
}
}


The inner loop scans i for the binary pattern 01 by iteratively shifting x to the right as long as it has 3 bits left. printf returns the number of characters printed, hence never 0, so the inner loop test fails after the printf, avoiding the need for a break statement.

# C++, 129 128 bytes

Adapting the same idea, the C++ variant is here:

#include<iostream>
long i,x;int main(int a,char**b){for(;++i<1L<<atoi(b);)for(x=i;x>4&&(x%4!=1||!(std::cout<<i<<','));x/=2);}


Technically, I should make i a long long to ensure 64 bit operation and compute upto 2^32 for an extra 5 bytes, but modern platforms have 64 bit ints.

# JavaScript ES6, 60 bytes

## Code

n=>[...Array(1<<n).keys()].filter(g=x=>x>4?x%4==1|g(x>>1):0)


Try it online!

## Explanation

[...Array(1<<n).keys()]                                          Create an array of numbers from 0 to 2^n-1
.filter(                                  Find the numbers which meet criteria
g=x=>x>4?                  :0     If x is less than or equal to four return zero (false/invalid)
x>4?x%4==1|              If the binary pattern ends with 01 return one (true/valid)
g(x>>1)       Otherwise bitshift right by one and try again


# PowerShell, 80 bytes

while([Math]::Pow(2,$args)-gt$n++){$n|?{[Convert]::ToString($n,2)-match"01"}}


# Python, 44 Bytes

Okay, this could probably be shorter but it's my first codegolf:

x=1
while True:
if '01' in bin(x):
print(x)
x+=1


Think this answers the question, please don't down vote if it doesn't, just post whats wrong with it below.

• You need to take input (input() is ideal) to get n, and then only count up to 2^n-1, rather than looping indefinitely. Also, you can use 1 and 2 spaces for the nesting, rather than 4 and 8, and using map or a list comprehension would probably shorten your code tremendously.
– user45941
Oct 23, 2015 at 13:50

another different matlab answer of good score.

## matlab 60(57)

@(n)find(mod(log2(bitcmp(1:2^n,fix(log2(1:2^n)+1))+1),1))


## execution

>> @(n)find(mod(log2(bitcmp(1:2^n,fix(log2(1:2^n)+1))+1),1))


ans =

@(n)find(mod(log2(bitcmp(1:2^n,fix(log2(1:2^n)+1))+1),1))


>> ans(5)

ans =

 5     9    10    11    13    17    18    19    20    21    22    23    25    26    27    29


• The idea is picking up numbers x where the binary representation of -(x)+1 doesnt contain only one occurence of 1

example:

0000111 is rejected because ~x=1111,~x+1=00001 contains one digit=1

0100111 is accepted because ~x=1011,~x+1=0111 contains many 1's

# k4, 32 28 bytes

{&1<+/'1='-':'0b\:'o:!*/x#2}


explanation

                      */x#2   /x^2
o:!        /assign o to [0,x^2 -1), e.g. for x=2, o<-0 1 2 3
0b\:'           /get binary value of each array elem
-':'                /get delta of adjacent digits in each binary num
+/'1='                    /sum each array to find how many deltas=1
&1<                          /return elems where there were greater than 1 deltas=1